How to calculate area of triangle with perpendicular lines
- In the figure below, the lines have slopes of 3 and 5. The lines intersect at (10,15). How far is it between the x-intercepts of the lines?
The equation of the line with slope of 3 is $y=3x+b$ and with the point $(10,15)$ is $15=3(10)+b$ so the equation is $y=3x-15$
Similarly, the slope is 5. So the equation is $y=5x-35$
The x intercepts are $x=5$ for $y=3x-15$ and $x=7$ for $y=5x-35$. The distance between the two 2. Is this correct?
- In the figure below, the two lines are perpendicular, and they intersect at $(6,8)$. The y-intercepts of the lines have a sum of zero. Find the area of the shaded region.
the equations of the lines are $y=mx+b$ and $y=-(\frac{1}{m})x-b$ since the y-intercepts are opposite each other. If I plug the point back in, I have:
$8=6m+b$ and $8=-(\frac{1}{m})(6)-b$
If I set these two equations together, I'd have
$6m+b=-(\frac{1}{m})(6)-b$
$6m=-\frac{6}{m}$
$6m^2 = -6$
$6m^2+6=0$
$6(m^2+1)=0$
Here I am stuck because I have a negative root so I'm pretty sure I attempted this problem wrong.
$\endgroup$ 31 Answer
$\begingroup$Your answer to the first question is correct.
For the second question, your error, as pointed out in the comment above, was that you accidentally cancelled the $b$'s, even though they had opposite signs. You have two equations:
$$8 = 6m+b \tag{1}$$
$$8 = -\left(\frac{1}{m}\right)6-b \tag{2}$$
There are multiple ways this system can be solved. For instance, you can isolate $b$ in equation $(1)$ and plug it in equation $(2)$.
$$8 = 6m+b \iff \color{blue}{b = 8-6m} \tag{1}$$
$$\implies 8 = -\left(\frac{1}{m}\right)6-\color{blue}{(8-6m)}$$
Can you take it on from here?
$\endgroup$ 5
